Does this property characterize abelian groups?












8












$begingroup$


Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?



This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.



If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.






abstract-algebra group-theory permutations







share|cite|improve this question











$endgroup$

















    8












    $begingroup$


    Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?



    This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.



    If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.






    abstract-algebra group-theory permutations







    share|cite|improve this question











    $endgroup$















      8












      8








      8


      2



      $begingroup$


      Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?



      This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.



      If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.






      abstract-algebra group-theory permutations







      share|cite|improve this question











      $endgroup$




      Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?



      This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.



      If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.






      abstract-algebra group-theory permutations




      abstract-algebra group-theory permutations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 2 at 23:29







      MathematicsStudent1122

















      asked Jan 2 at 22:44









      MathematicsStudent1122MathematicsStudent1122

      8,67622467




      8,67622467






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Yes. By induction on $kge 2$; $k=2$ is clear.



          Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.



          Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060064%2fdoes-this-property-characterize-abelian-groups%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Yes. By induction on $kge 2$; $k=2$ is clear.



            Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.



            Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Yes. By induction on $kge 2$; $k=2$ is clear.



              Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.



              Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Yes. By induction on $kge 2$; $k=2$ is clear.



                Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.



                Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.






                share|cite|improve this answer









                $endgroup$



                Yes. By induction on $kge 2$; $k=2$ is clear.



                Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.



                Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 2 at 23:32









                YCorYCor

                7,808929




                7,808929






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060064%2fdoes-this-property-characterize-abelian-groups%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Ellipse (mathématiques)

                    Image
                    Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                    Read more

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                    Read more